Answer :
Answer:
S_x = -(2/3)*t^3 + 4*t
S_x = 3.77 m
a_x = -4*t
Explanation:
Given:
- The velocity of the object as a function of time t is given:
V_x = a - B*t^2
- Where, a = 4.0 m/s ; B = 2.0 m/s^3
- At time t = 0, x = 0
Find:
(a) Calculate the object’s position and acceleration as functions of time.
(b) What is the object’s maximum positive displacement from the origin?
(c) What is the object's acceleration as a function of time?
Solution:
- We will use differential calculus to solve this problem, we know that velocity is the derivative of the displacement:
V_x = ds_x / dt
(4.0-2*t^2) = ds_x / dt
- Separate variables:
(4.0-2*t^2). dt = ds_x
- Integrate both sides with t = 0, s_x = 0:
4.0*t - (2/3)t^3 = s_x
S_x = -(2/3)*t^3 + 4*t
- Acceleration is the derivative of velocity, so we will differentiate the given expression as follows:
a_x = dv_x / dt
a_x = d (4.0-2*t^2) / dt
- Differential v_x with respect to x:
a_x = -4*t
- The maximum displacement from starting point can be calculated by setting v_x to zero:
V_x = 0
(4.0-2*t^2) = 0
t^2 = 2
t = sqrt (2) = 1.4142 s
- Evaluate S_x at t = sqrt(2)
S_x = -(2/3)*2^(3/2) + 4*2^0.5
S_x = 3.77 m